now, there are 12 months in a year, so 18 months is really 18/12 of a year, thus
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$4000\\ P=\textit{original amount deposited}\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ t=years\to \frac{18}{12}\dotfill &\frac{3}{2} \end{cases} \\\\\\ 4000=P[1+(0.05)(\frac{3}{2})]\implies 4000=P(1.075) \\\\\\ \cfrac{4000}{1.075}=P\implies 3720.93\approx P](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5Cdotfill%20%26%20%5C%244000%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5C%5C%20r%3Drate%5Cto%205%5C%25%5Cto%20%5Cfrac%7B5%7D%7B100%7D%5Cdotfill%20%260.05%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B18%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%204000%3DP%5B1%2B%280.05%29%28%5Cfrac%7B3%7D%7B2%7D%29%5D%5Cimplies%204000%3DP%281.075%29%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B4000%7D%7B1.075%7D%3DP%5Cimplies%203720.93%5Capprox%20P)
The given sample's confidence interval is 4.73 ± 0.1199, or from 4.61 to 4.85
<h3>What is the z score?</h3>
The z-score is a numerical assessment of a value's connection to the mean of a set of values, expressed in terms of standards from the mean, that is used in statistics.
Given data;
Mean = 4.73
Standard deviation = 0.865
Sample size = 200
interval = 95%
Confidence interval=?
95% of samples contain the population mean (μ) within the confidence interval of 4.73 ± 0.1199.
Hence, the sample's confidence interval is 4.73 ± 0.1199, from 4.61 to 4.85
To learn more about the Z score, refer to;
brainly.com/question/15016913
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Answer:
I don't seem to be so sure, but according to what i did the answer is A